Why is the entropy change of surroundings in free expansion zero?

Question

I have read that what differentiates an irreversible process like adiabatic free expansion from its counterpart reversible process — isothermal expansion, which takes the system to the same final state — is the change in entropy of the surroundings. In the free expansion case it is $0,$ whereas in isothermal expansion $\displaystyle \Delta S_\mathrm{surr} = -nR\ln\frac{V_2}{V_1}$ such that $\Delta S_\mathrm{tot} = 0$.

I am unable to understand why the entropy change of the surroundings in the free expansion case is $0$.

$$\Delta S_\mathrm{surr} = \int \frac{\mathrm dq_\mathrm{irrev}}{T_\mathrm b} + S_\mathrm{gen} = \int \frac{\mathrm dq_\mathrm{rev}}{T}$$

Since the entropy transfer term $\displaystyle \int \frac{\mathrm dq_\mathrm{irrev}}{T_\mathrm b}$ is $0$ as the process is adiabatic, for the net entropy to equate to $0$ the entropy generated must also be $0.$ But doesn't this imply that the surrounding is undergoing a reversible process?

Also, I am unable to think of a suitable reversible process for the surroundings which would take it to the same final state. Moreover, what does the change in entropy even signify for a vacuum?

Answer 1

Since entropy is a state variable, $\Delta S_\mathrm{sys}$ between two states (defined by variables such as T, p, V, n) is the same whether the process is carried out reversibly or irreversibly. According to the second law of thermodynamics, for the system

$$\Delta S_\mathrm{sys} = \int \frac{dq_{rev}}{T} $$

Again, the value $\Delta S_\mathrm{sys}$ will be the same for an irreversible process even though the heat exchanged during the process will be $\int dq$, not $\int dq_{rev}$.

For the surroundings $$\Delta S_\mathrm{surr} = -\int \frac{dq}{T}$$

If the surroundings consists of vacuum, its temperature is nominally absolute zero and entropy is zero (here according to the third law of thermodynamics).